Unit 2 Multiple Choice Review

1. The mean number of points per game scored by basketball players during a high school championship is 9.4, and the standard deviation is 1.5. Assuming that the number of points are normally distributed, what number of points per game will place a player in the top 15% players taking part in the basketball championship?

A.) 8.54 points per game

B.) 9.93 points per game

C.) 10.95 points per game

D.) 12.56 points per game

E.) 13.91 points per game

#1 - Solution C - The player scores more points per game than 85% of

competitors, so from the table of standard normal

probabilities, z = 1.037.

1.037 = (x - μ)/σ

1.037 = (x - 9.4)/1.5

x = 10.95 points/game.

2. The heights of 100 students are normally distributed with a mean of 172cm. What is the standard deviation of the heights given that the probability of a height above 180cm is .25? A.) 7.7cm

B.) 8cm

C.) 9.2cm

D.) 10cm

E.) 11.9cm

#2 - Solution E –

P(height > 180cm) = .25

P(height < 180cm) = .75

P(z < (180 - 172)/σ) = .75

(180 - 172)/σ = .675

σ = 11.9cm

3. Two distributions D1 and D2 are displayed on the same graph. If the distribution D1 is right skewed, the distribution D2 is left skewed, and the mean of D1 is lower than the mean of D2, which of the following statements is not true? A.) The median of D1 is lower than the median of D2

B.) The mean of D1 is lower than the median of D2.

C.) The mean of D2 is lower than the median of D1.

D.) The median of D1 is lower than the mean of D2.

E.) The mean of D2 is lower than the median of D2.

#3 - Solution C – The mean of D2 > mean of D1, and the median

of D1 must be less than the mean of D1 because the data is right skewed, so the mean of D2 cannot be less than the median of D1

D1 Med D1 Mean D2 Mean D2 Med

D1 D2

4. If the heights of 99.7% of American men are between 5’0” and 7’0”, what is your estimate of the standard deviation of the height of American men? (Assume the heights of American Men are Normally Distributed).

A.) 1"

B.) 2"

C.) 4"

D.) 6"

E.) 12"

#4 - Solution C – 4”

First convert to inches.

By the Empirical Rule, we know that 99.7% of all normally distributed data lie within 3 standard deviations of the mean – so that is 6 standard deviations between 60” and 84”.

5 '0" 60" and 7 '0" 84"

84 60

64

5. In the distribution shown below, five of the points make up the 5 number summary. Which 2 points are NOT part of the 5 number summary for this distribution?

A.) B & E

B.) C & F

C.) C & E

D.) B & F

E.) A & C

#6 - Solution D - B & F

6. In May, 316 students took the AP Economics exam. The exam scores were approximately normally distributed. The five number summary of the scores is: 0 26 31 36 50 Therefore the Variance of test scores is approximately A.) 5

B.) 7

C.) 18

D.) 56

E.) 54

#6 - Solution D - If the data is normally distributed, then the mean is

approximately equal to the median. Using the Standard Normal Table (Table A):

2 2

     0.25,  0.67

26 310.67   

7.46269

(7.46269) 55.69

X

X

X

where p z

7. In a given set of data, the median is equal to the mean. Therefore, we can conclude: A.) The distribution is approximately Normally Distributed.

B.) The distribution is skewed.

C.) The distribution is symmetrical.

D.) The distribution is bell-shaped.

E.) We cannot conclude anything about the shape of the distribution with any certainty.

#7 - Solution E – We can’t conclude anything about the shape of the

distribution with any certainty.

(There is not enough information to make a decision).

8. The Normal Probability Plot shown below indicates that the data set is A.) Strongly skewed right.

B.) Strongly skewed left.

C.) Approximately Normal.

D.) Approximately Symmetrical.

E.) We cannot conclude the shape of the distribution from this plot.

#8 - Solution A – The plot suggests that the data is skewed heavily to

the right.

9. The density curve takes the value of 0.10 for and 0 for all other values of x. What percent of x values lie between 3 and 7?

A.) 20%

B.) 30%

C.) 40%

D.) 50%

E.) 60%

1 11x

1 11 0

1

10

#9 - Solution C – 40%

7 – 3 = 4

4(1/10) = .40 or 40%

1 11 0 3 7

1

10

10. When answering any free-response problem in AP Statistics regarding Normal Distributions, one should always A.) State the problem in terms of x and justify why normal

calculations are appropriate.

B.) Draw, label and shade the Normal Curve.

C.) Write the problem in terms of x then show and justify the calculations.

D.) Clearly state a conclusion in context of the problem.

E.) All of the above!

E – ALL of the above ALWAYS!